Question

Tìm x

a) \(\left|x-7\right|+2x+9=6\)

b)  \(2.\left|2x-3\right|=\frac{1}{2}\)

c) \(\left|3x-4\right|+\left|5y+5\right|=0\)

Answer 1

b, 2./2x-3/=\(\frac{1}{2}\)

=> /2x-3/=\(\frac{1}{4}\)

=> \(\orbr{\begin{cases}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{cases}}\)=>\(\orbr{\begin{cases}2x=\frac{13}{4}\\2x=\frac{11}{4}\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{13}{8}\\x=\frac{11}{8}\end{cases}}\)

vậy x=\(\frac{13}{8}\)hoặc x= \(\frac{11}{8}\)

c,/3x-4/+/5y+5/=0. Do /3x-4/\(\ge0\),/5x+5/\(\ge0\)

=>\(\hept{\begin{cases}3x-4=0\\5y+5=0\end{cases}}\)=>\(\hept{\begin{cases}3x=4\\5y=-5\end{cases}}\)=>\(\hept{\begin{cases}x=\frac{4}{3}\\y=-1\end{cases}}\)

vậy ................................-_-???????!!!!! @-@

còn phần a thì mình ko biết